What would (2,2) be as a Equation??

50 pts please help me due today links or incorrect answers will get reported

Bumek [7]

Using the normal distribution, the area underneath the shaded region <u>between the two z-scores</u> is given by:

C. 0.6766.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

Hence, for this problem, the area is the <u>p-value of z = 0.75 subtracted by the p-value of z = -1.3</u>.

Looking at the z-table, the p-values are given as follows:

z = 0.75: 0.7734.

z = -1.3: 0.0968.

Then:

0.7734 - 0.0968 = 0.6766.

Which means that option C is correct.

More can be learned about the normal distribution at brainly.com/question/15181104

#SPJ1

3 0

2 years ago

Please determine whether the set S = x^2 + 3x + 1, 2x^2 + x - 1, 4.c is a basis for P2. Please explain and show all work. It is

ohaa [14]

The vectors in $S$ form a basis of $P_2$ if they are mutually linearly independent and span $P_2$ .

To check for independence, we can compute the Wronskian determinant:

$\begin{vmatrix}x^2+3x+1&2x^2+x-1&4\\2x+3&4x+1&0\\2&4&0\end{vmatrix}=4\begin{vmatrix}2x+3&4x+1\\2&4\end{vmatrix}=40\neq0$

The determinant is non-zero, so the vectors are indeed independent.

To check if they span $P_2$ , you need to show that any vector in $P_2$ can be expressed as a linear combination of the vectors in $S$ . We can write an arbitrary vector in $P_2$ as

$p=ax^2+bx+c$

Then we need to show that there is always some choice of scalars $k_1,k_2,k_3$ such that

$k_1(x^2+3x+1)+k_2(2x^2+x-1)+k_34=p$

This is equivalent to solving

$(k_1+2k_2)x^2+(3k_1+k_2)x+(k_1-k_2+4k_3)=ax^2+bx+c$

or the system (in matrix form)

$\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}$

This has a solution if the coefficient matrix on the left is invertible. It is, because

$\begin{vmatrix}1&1&0\\3&1&0\\1&-1&4\end{vmatrix}=4\begin{vmatrix}1&2\\3&1\end{vmatrix}=-20\neq0$

(that is, the coefficient matrix is not singular, so an inverse exists)

Compute the inverse any way you like; you should get

$\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}=-\dfrac1{20}\begin{bmatrix}4&-8&0\\-12&4&0\\-4&3&-5\end{bmatrix}$

Then

$\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}\begin{bmatrix}a\\b\\c\end{bmatrix}$

$\implies k_1=\dfrac{2b-a}5,k_2=\dfrac{3a-b}5,k_3=\dfrac{4a-3b+5c}{20}$

A solution exists for any choice of $a,b,c$ , so the vectors in $S$ indeed span $P_2$ .

The vectors in $S$ are independent and span $P_2$ , so $S$ forms a basis of $P_2$ .

5 0

3 years ago

Graph<br><br> F(x)=4x-3<br><br><br> Can someone help me?

AveGali [126]

Y = 4x - 3
y = mx + b
m = 4
b = -3
y intercept = (0,3)
(x,y) = (0,3)
(x,y) = (1,1) - substitute in ori equation

3 0

3 years ago

What is (are) the solution(s) to the equation 23x−3=−13x−6?

solniwko [45]

From Question ,
23x-3 = 13x-6
Or, 23x-13x = -6+3
Or, 10x = -3
Or, x = -3/10
Therefore x = -3/10 ans

Hope this helps !

Please mark it as the brainliest

6 0

3 years ago

Express the statement as an algebraic equation. Two thirds of a number is 89

Darya [45]

Answer:

The number is 133.5

Step-by-step explanation:

Let the number = x

2/3 * x = 89 Multiply by 3

2x = 89 * 3 Combine

2x = 267 Divide by 2

x = 133.5

8 0

3 years ago